# Physics 9702 Doubts | Help Page 116

__Question 588: [Temperature > Internal energy]__**(a)**On Fig.1, place tick (✓) against those changes where internal energy of the body is increasing.

water freezing at constant
temperature ..........................................

a stone falling under gravity in a
vacuum ..........................................

water evaporating at constant
temperature ..........................................

stretching a wire at constant
temperature ..........................................

**(b)**A jeweller wishes to harden a sample of pure gold by mixing it with some silver so that mixture contains 5.0% silver by weight. Jeweller melts some pure gold and then adds the correct weight of silver. Initial temperature of the silver is 27 °C. Use data of Fig.2 to calculate initial temperature of the pure gold so that the final mixture is at the melting point of pure gold.

gold silver

melting point / K 1340 1240

specific heat capacity (solid or
liquid) / J kg

^{–1}K^{–1}129 235
specific latent heat of fusion / kJ
kg

^{–1}628 105**(c)**Suggest a suitable thermometer for measurement of the initial temperature of the gold in (b).

**Reference:**

*Past Exam Paper – June 2003 Paper 4 Q2*

__Solution 588:__**(a)**

water freezing at constant
temperature ..........................................

a stone falling under gravity in a
vacuum ..........................................

water evaporating at constant
temperature tick

stretching a wire at constant
temperature tick

{

__The concept of ‘Internal Energy’__
Internal energy is the sum
of potential and kinetic energy due to the RANDOM motion of the molecules.

It is important to point
out that the internal energy is due to the energies of the randomly moving
molecules only AND as the word says internal energy is due to the ‘internal’
structure of the system (due to molecules or atoms), not the whole system
itself.

For example, for a car
moving with speed v or moving up a slope. The sum of its kinetic energy and
gain in potential energy do NOT give the internal energy. It only gives the
total energy of the car.

So, again when we are
talking about internal energy, we are firstly considering the atoms / molecules
inside the materials and secondly, we are only considering those atoms / molecules
that are in random motion.}

{Case 1: water freezing at
constant temperature

For constant temperature,
the kinetic energy of the molecules is unchanged. For freezing, there is a
change in phase from liquid to solid occurring. This requires energy to form
the new bonds. So, the material is actually losing some of its potential energy
to form the bonds. So, internal energy is decreasing in this case.

Case 2: a stone falling
under gravity in a vacuum

This case is dealing with the
energy of a large object, not its internal structure. As explained before, in
such a case, we do not talk about internal energy, but rather the total energy
of the stone.

Case 3: water evaporating
at constant temperature

Again for constant
temperature, the kinetic energy of the molecules is unchanged. For evaporation,
there is a change in phase from liquid to gas occurring. The potential energy
between the randomly moving molecules as they go in the gas state (more
molecules are now in random motion). So, internal energy is increasing in this case.

Case 4: stretching a wire
at constant temperature

Again for constant
temperature, the kinetic energy of the molecules is unchanged. For stretching a
wire, work needs to be done on the wire. Since the wire has now become longer
than its initial length, the molecules / atoms inside the wire have more space
to move freely / randomly compared to its initial shape where all the molecules
/ atoms were packed together (in the solid wire). We say that the wire has
elastic potential energy. So, the internal energy of the wire increases.}

**(b)**

{To reach the final
temperature which is equal to the melting point of pure gold, the temperature
of the silver (initially at 300K) should increase. This means that the
temperature of the pure gold should have decreased to provide energy to the
silver.

Î”Q = mcÎ”Î¸

m is the total mass. 95%
of m is gold and 5% of m is silver.}

Heat lost by liquid gold = 0.95m × 129
× Î”T

{The melting point go gold
is higher than that of silver. So, before reaching the final temperature, the silver
has melted first. This requires energy which is given by Î”Q = mL = 0.05m × 105 000 where L is the specific latent heat of fusion.

After melting, the silver
needs to increase its temperature to the final temperature. This further requires
energy given by Î”Q = mcÎ”Î¸ = 0.05m × 235 × (1340 – 300)}

Heat gained (by silver) = 0.05m × 235
×
(1340 – 300) + 0.05m × 105 000

{Heat lost by liquid gold
= heat gained by silver}

122.5mÎ”T = 17 470m

Change in temperature Î”T = 143 K

{Final temperature is the
melting point of gold}

Initial temperature of gold = 143 +
1340 = 1483 K

**(c)**Example: thermocouple / resistance thermometer

__Question 589: [Current of Electricity]__
Four resistors of equal value are
connected as shown.

How will powers to the resistors
change when resistor W is removed?

A The powers to X, Y and Z will all
increase.

B The power to X will decrease and
the powers to Y and Z will increase.

C The power to X will increase and
the powers to Y and Z will decrease.

D The power to X will increase and
the powers to Y and Z will remain unaltered.

**Reference:**

*Past Exam Paper – June 2011 Paper 11 Q36*

__Solution 589:__**Answer: C.**

Power dissipated in a resistor = I

^{2}R
Connecting 2 resistors of, say,
resistance R in series causes the overall resistance to increase (= R+R = 2R).

Connecting 2 resistors of, say,
resistance R in parallel causes the overall resistance to decrease (= [1/R +
1/R]

^{-1}= 0.5R).
Since the 4 resistors have the same
resistance in the connection shown, the p.d. across each parallel connection
(and each resistor) will be the same.

Overall resistance in circuit = 0.5R
+ 0.5R = R

Let the e.m.f. in the circuit = E

Current in the circuit = E / R = I

The total current splits at the
parallel combination.

Power dissipated in X = (I/2)

^{2}R = I^{2}R / 4 = 0.25 I^{2}R
The same power is dissipated in all
the resistors.

When W is removed, the overall
resistance of the parallel combination of W and X changes from 0.5R to R (due
to X only).

Total resistance in new circuit = R
+ 0.5R = 1.5R

Current in circuit = E / 1.5R =
(2/3)I

The overall current in the circuit
will also change – it decreases since the overall resistance in the complete
circuit has increased. This current flows through X but is split into half at
the combination of Y and Z. So, the current through Y and Z is half that
through X.

Power dissipated = I

^{2}R
Power dissipated in X = [(2/3)I]

^{2}R = 4 I^{2}R / 9 = 0.44 I^{2}R
Half the current flows through Y and
Z. The same power is dissipated through both resistors.

Power dissipated in Y = [0.5 (2/3)I]

^{2}R = I^{2}R / 9 = 0.11 I^{2}R
So, the power in X increases and the
power in Y (and Z) decreases.

__Question 590: [Electromagnetism > Hall Probe]__**(a)**Constant current is maintained in a long straight vertical wire. Hall probe is positioned a distance r from the centre of the wire, as shown in Fig.1.

(i) Explain why, when Hall probe is
rotated about the horizontal axis XY, the Hall voltage varies between a maximum
positive value and a maximum negative value.

(ii) Maximum Hall voltage V

_{H}is measured at different distances r.
Data for V

_{H}and the corresponding values of r are shown in Fig.2.
V

_{H}/ V r / cm
0.290 1.0

0.190 1.5

0.140 2.0

0.097 3.0

0.073 4.0

0.060 5.0

It is thought that V

_{H}and r are related by expression of the form
V

_{H}= k / r
where k is a constant.

1. Without drawing a graph, use data
from Fig.2 to suggest whether the expression is valid.

2. A graph showing variation with 1
/ r of V

_{H}is plotted.
State features of the graph that
suggest that the expression is valid.

**(b)**Hall probe in (a) is now replaced with a small coil of wire connected to sensitive voltmeter. Coil is arranged so that its plane is normal to the magnetic field of the wire.

(i) State Faraday’s law of
electromagnetic induction and hence explain why voltmeter indicates a zero
reading.

(ii) State three different ways in
which an e.m.f. may be induced in the coil.

**Reference:**

*Past Exam Paper – June 2010 Paper 41 Q5*

__Solution 590:__**(a)**

(i) The Hall voltage V

_{H}depends on the angle between (the plane of) the probe and B-field (magnetic field).
either The Hall voltage V

_{H}is maximum when the plane and the B-field are normal to each other.
or The Hall voltage V

_{H}is zero when the plane and the B-field are parallel
or The Hall voltage V

_{H}depends on the sine of the angle between the plane and B-field
(ii)

1.

{From the formula, the
product of VH and r is a constant. k = V

_{H }r}
Calculate V

_{H}r at least three times {any sets can be chosen}
1

^{st}set: 0.29 × 1.0 = 0.29 (to 2sf) = 0.3
2

^{nd}set: 0.19 × 1.5 = 0.285 = 0.29 (to 2sf) = 0.3
3

^{rd}set: 0.14 × 2.0 = 0.28 (to 2sf) = 0.3
EITHER If given to 1 s.f., the
product is constant, so the expression is valid or approximately constant so
valid

OR If given to 2 s.f., the product
is not constant, so the expression is invalid

2. The graph is a straight line that
passes through the origin

**(b)**

(i) Faraday’s law of electromagnetic
induction states that the e.m.f. induced is proportional / equal to the rate of
change of (magnetic) flux (linkage).

The voltmeter indicates a zero
reading because there is a constant field in

__coil__/ flux (linkage) of__coil__does not change
(ii) Example:

Vary the current (in wire) / switch
current on or off / use a.c. current

Rotate the coil

Move the coil towards / away from
wire

__Question 591: [Work, Energy and Power]__
Box of weight 30 N is released from
rest on a ramp that is at angle of 30° to the horizontal. The box slides down
the ramp so that it falls through a vertical distance of 8.0 m. A constant frictional
force of 10 N acts on box while it is moving.

What is the kinetic energy of box
after falling through this distance?

A 80 J B 160 J C
240 J D 400 J

**Reference:**

*Past Exam Paper – November 2014 Paper 13 Q17*

__Solution 591:__**Answer: A.**

In this question, there is a
frictional force of 10N that opposes the motion of the box down the ramp. So,
work needs to be done against friction.

Consider a case where the ramp was
frictionless. From the conservation of energy, the gravitational potential
energy would be converted into kinetic energy of the box as it falls.

But in this case, friction is
present. So, not all of the gravitational potential energy would be kinetic
energy. Some of the energy is used up as work done against friction.

Loss in GPE = Gain in KE + Work
against friction

Loss in GPE = mgh = (mg)h = (30) (8)
= 240 J

Let the distance travelled by the
box along the ramp = s

sin30 = 8 / s

Distance s = 8 / sin30 = 16m

Work done against friction = Fs = 10
(16) = 160 J

Loss in GPE = Gain in KE + Work
against friction

240 = Gain in KE + 160

Gain in KE = 240 – 160 = 80 J

Dear Sir, please consider answering these questions.

ReplyDelete01/O/N/07 Q.40

02/O/N/08 Q.7(b)(ii)

21/O/N/09 Q.6(b)

12/M/J/10 Q.6

21/M/J/10 Q.3(b)(iii)part1

22/M/J/10 Q.4(b)(ii),(c), Q.5(a)(ii), Q.7(b)(ii)2.

23/M/J/10 Q.7(b)

11/O/N/10 Q.1,24,31

12/O/N/10 Q.26,33

21/O/N/10 Q.3(c)(ii),Q.7(b)(ii)2.

22/O/N/10 Q.4(b)(ii)

11/M/J/11 Q.9,10,14,16,23,24,25,27,32,34

21/M/J/11 Q.7(b)

Your cooperation is highly appreciated.

For 01/O/N/07 Q.40, check question 598 at

Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-118.html

Due to the large number of questions in the post above, please consider answering (in details) only paper 2 questions as these are the first priority. You may answer the rest of the questions after one month from now. Thanks

DeleteFor 02/O/N/08 Q.7(b)(ii), check at

Deletehttp://physics-ref.blogspot.com/2014/09/9702-november-2008-paper-2-worked.html

Sir, please consider answering the rest of the paper 2 by now.

DeleteFor 21/O/N/09 Q.6(b), check solution 612 at

Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-121.html

This comment has been removed by the author.

DeleteFor 22/M/J/10 Q.4(b)(ii),(c), check at

Deletehttp://physics-ref.blogspot.com/2014/10/9702-june-2010-paper-22-worked.html

Hello..can you help me with questions from oct/nov 2009 paper 12 questions 8,9,11,12, 25 and 27

ReplyDeleteThank you

Check if they are explained at

Deletehttp://physics-ref.blogspot.com/2014/11/9702-november-2009-paper-12-worked.html

if not, ask the specific questions there

thank you..i checked and found all explanation but not Q9

Deletekindly help me with Q9

Dear Sir, please consider answering these questions.

ReplyDelete02/O/N/08 Q.7(b)(ii)

21/O/N/09 Q.6(b)

21/M/J/10 Q.3(b)(iii)part1

22/M/J/10 Q.4(b)(ii),(c), Q.5(a)(ii), Q.7(b)(ii)2.

23/M/J/10 Q.7(b)

21/O/N/10 Q.3(c)(ii),Q.7(b)(ii)2.

22/O/N/10 Q.4(b)(ii)

21/M/J/11 Q.7(b)

See the links in the comments above

DeleteHere are the remaining questions for now:

ReplyDelete21/M/J/10 Q.3(b)(iii) part 1

22/M/J/10 Q.5(a)(ii), Q.7(b)(ii)2.

23/M/J/10 Q.7(b)

21/O/N/10 Q.3(c)(ii),Q.7(b)(ii)2.

22/O/N/10 Q.4(b)(ii)

For 21/M/J/10 Q.3(b)(iii) part 1

Deletehttp://physics-ref.blogspot.com/2014/10/9702-june-2010-paper-21-worked.html